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Quarter-turn rotation is not diagonalisable over $R$ but is diagonalisable over $C$

Example

The quarter-turn matrix

$R = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$

is not diagonalisable over $R$ , but it is diagonalisable over $C$ .

Facts & Assumptions

Given: The displayed real matrix $R$ .

[L1]

An endomorphism is diagonalisable exactly when its minimal polynomial is a product of distinct linear factors over the base field (An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors).

[L2]

The polynomial $x^{2} + 1$ is irreducible over $R$ ( $x^{2} + 1$ is irreducible over $R$ ).

[L3]

In $C$ , the element $i$ satisfies $i^{2} = - 1$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ).

Verification

technique · direct

1.1

Direct multiplication gives $R^{2} = - I$ , while no linear polynomial annihilates $R$ , so $μ_{R} = x^{2} + 1$ . By [L2] and [L1], $R$ is not diagonalisable over $R$ .

L1L2algebra

2.1

Over $C$ , [L3] gives $x^{2} + 1 = (x - i) (x + i)$ with distinct roots. By [L1] the extended matrix is diagonalisable; explicitly $(i, 1)$ and $(- i, 1)$ are eigenvectors for $i$ and $- i$ and form a basis.

L1L3algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 52 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Keith Conrad, Potential Diagonalizability, Example 1 (standard reference, not scraped)

Quarter-turn rotation is not diagonalisable over RR but is diagonalisable over CC